如何更新 dplyr 管道中的值？答案

【问题标题】：How to update values in a dplyr pipe?如何更新 dplyr 管道中的值？
【发布时间】：2020-06-09 19:52:46
【问题描述】：

我想更新新列中的值。

这是我的数据：

people<- c("father", "parents", "father", "children", "girl", "boy", "grand father", "grand mother", "grandparents" ) 
dataset0 <- data.frame(people)
dataset0

然后输出：

father              
parents             
father              
children                
girl                
boy             
grand father                
grand mother                
grandparents

预期输出：

 people           people_update

father            parents   
parents           parents   
father            parents   
children          children
girl              children
boy               children
grand father      grandparents          
grand mother      grandparents      
grandparents      grandparents

我尝试使用replace()

dataset <- dataset0 %>%
   mutate(people_update = replace(people, people =="girl", "children")) %>%
   mutate(people_update = replace(people, people =="boy", "children")) 
 dataset

但这不起作用。第二个mutate() 命令取消了第一个mutate() 命令。

【问题讨论】：

标签： r dplyr

【解决方案1】：

尝试case_when 指定多个替换。它比多个ifelse 语句更简洁。

library(dplyr)

dataset <- dataset0 %>%
  mutate(people_update = case_when(
    people %in% c("father", "parents")                            ~ "parents",
    people %in% c("children", "girl", "boy")                      ~ "children",
    people %in% c("grandparents", "grand father", "grand mother") ~ "grandparents",
    TRUE                                                          ~ NA_character_
  ))

【讨论】：

0.5 版本。我只有人员列。
我的是0.7.1。这个case_when 功能是新事物。请更新到最新版本。
如果还是不行，使用多个ifelse语句或者加入其他人建议的方法。

【解决方案2】：

这可以通过嵌套的ifelse 语句来处理，即

library(dplyr)

dataset0 %>% 
  mutate(v1 = ifelse(people %in% c('father', 'mother', 'parents'), 'parents', 
             ifelse(people %in% c('girl', 'boy', 'children'), 'children', 'grandparents')))

#        people           v1
#1       father      parents
#2      parents      parents
#3       father      parents
#4     children     children
#5         girl     children
#6          boy     children
#7 grand father grandparents
#8 grand mother grandparents
#9 grandparents grandparents

【讨论】：

【解决方案3】：

case_when() 或嵌套if_else() 的替代方法是加入转换表map：

library(dplyr)
dataset0 %>% left_join(map)

Joining, by = "people"
        people people_update
1       father       parents
2      parents       parents
3       father       parents
4     children      children
5         girl      children
6          boy      children
7 grand father  grandparents
8 grand mother  grandparents
9 grandparents  grandparents
Warning message:
Column `people` joining factor and character vector, coercing into character vector

map 由下式给出

map <- tribble(
  ~people, ~people_update,
  "father",            "parents",   
  "parents",           "parents",   
  "children",          "children",
  "girl",              "children",
  "boy",               "children",
  "grand father",      "grandparents",          
  "grand mother",      "grandparents",      
  "grandparents",      "grandparents"
)
map

# A tibble: 8 x 2
        people people_update
         <chr>         <chr>
1       father       parents
2      parents       parents
3     children      children
4         girl      children
5          boy      children
6 grand father  grandparents
7 grand mother  grandparents
8 grandparents  grandparents

如果需要翻译的项目很少，可以修改代码：

# define only items to be changed
map2 <- tribble(
  ~people, ~people_update,
  "father",            "parents",   
  "mother",            "parents",   
  "girl",              "children",
  "boy",               "children",
  "grand father",      "grandparents",          
  "grand mother",      "grandparents"      
)

请注意，"mother" 已添加到转换表中。

dataset0 %>% 
  left_join(map2) %>% 
  # copy unchanged items
  mutate(people_update = if_else(is.na(people_update), people, people_update))

        people people_update
1       father       parents
2      parents       parents
3       father       parents
4     children      children
5         girl      children
6          boy      children
7 grand father  grandparents
8 grand mother  grandparents
9 grandparents  grandparents

【讨论】：

【解决方案4】：

重点是你的第二个mutate，你在x参数处一直使用people而不是people_update。

dataset <- dataset0 %>%
               mutate(people_update = replace(people, people == "girl", "children")) %>%
               mutate(people_update = replace(people_update, people == "boy", "children")) 
dataset

【讨论】：